![]() This is not really an answer to the original question, both because this proof is in the triangulated setting and does not avoid the isomorphism of the original question. (It took me a long time to thus actually understand the argument of Zeeman.) Out of frustration about this state of affairs I wrote up a variant of Zeeman's proof (based on a treatment by Thomassen), but including the well-definedness of attaching handles/cross-caps. Especially in the topological setting, this is a subtle point, requiring even in dimension 2 a kind of Schönflies theorem (and being a very difficult theorem in dimension 4). The second part is essentially equivalent to the well-definedness of the connected sum. The isomorphism type only depends on the number of handles and cross-caps attached.īoth the Zip-proof and the Zeeman proof refered to above in the comments only prove the first part and not the second part. But even if we choose such a setting, there are two statements one has to prove (at least in one approach):Įvery closed surface is isomorphic to a sphere with handles or cross-caps attached. Of course, there are the differentiable, triangulated and topological setting. The first thing to note is that there are different strenghts of the classification theorem for surfaces. This is more an extended comment than an answer to the question. The triangulability of surfaces is of course not so easy, its proof in Moise uses the Schoenflies theorem. (Uniformization has, besides the classical proof via the Dirichlet principle, by now more conceptual proofs via circle packing or via Ricci flow.)įinally, if you believe that every surface can be triangulated, then the classification of surfaces is proven by a not so difficult combinatorial inductive argument. Yet another approach would be to use uniformization and thus reduce the classification of surfaces to classification of discrete, torsion-free subgroups of PSL(2,R). (For this argument you need that every orientable surface is complex this follows from the homotopy equivalence $GL^ (2,R)\sim GL(1,C)$ and the obvious vanishing of the Nijenhuis tensor.) This can be seen from the Riemann-Roch theorem, which for $\chi(S)=2$ implies existence of a meromorphic function with only one pole of order 1, thus a biholomorphic map to $P^1C$. Thus the proof of classification boils down to show that a (closed, orientable) surface with $\chi(S)=2$ must be $S\cong S^2$. It is actually not hard to prove by purely topological arguments that for a (closed, orientable) surface S with $\chi(S)<2$ there is another surface S‘ with $\chi(S^\prime)=\chi(S) 2$. By induction, each component of $S^\prime$ is $S^2$ with finitely many handles added, so the same is true for $S$. The Morse function extends (constant on the disks). Consider the (possibly disconnected) surface $S^\prime$ obtained by cutting off the pair of pants and gluing in three disks. A neighborhood of a saddle point is a pair of pants, with the Morse function constant on boundary curves. For the inductive step, consider a Morse function with k saddle points. Assume by induction that a surface with k-1 saddle points is $S^2$ with finitely many handles added. If it has no saddle points, then (using the gradient flow) $S\cong S^2$. Take a Morse function on the (closed, orientable) surface S. Repeat for the other drain.I guess the most conceptual proof is the one using Morse theory: Now trim this boundary polyline to the intersection polyline you made in steps 1 - 5 and apply it as an outer boundary for the drain. ![]() This is now a polyline that you can use to trim your 2 drains to each other.įor each drain - select and choose extract from surface, extract object. You should only have 1 choice - major contour. Select this contour/surface and choose "extract from surface", extract objects from the ribbon. On the display tab make sure only major contour is shownĥ)Once you do this the intersection between the surfaces should show up as a surface with 1 contour. On the contours tab change the intervals to something huge like 10000 minor, 25000 major. It doesn't matter which one you pick first.ģ) This will result in a new surface that is a volume comparison surface, but if you change the way it displays you can make it show the intersection between the drainsĤ)Right click this surface in the prospector tab of toolspace. It will ask you to select a TOP and BOTTOM surface. 1) Make sure your drains overlap/intersect - it's ok if they overlap by more than is necessaryĢ) type createsurfacecomposite into the command line.
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